A Short Problem that Leads to?

In \(\triangle ABC\), \(\overline{BC}=a,\overline{CA}=b,\overline{AB}=c\)

PP is a point in ABC\triangle ABC. Use a,b,ca,b,c to show the minimum value of PA+PB+PC\overline{PA}+\overline{PB}+\overline{PC}

#Geometry

Note by X X
2 years, 8 months ago

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Comments

P is actually the Fermat Point of the triangle!!! So, after a bit of calculations, we get the final answer as shown in the link below............( I'm sorry since I don't know LATEX I have just input the answer in Desmos.......)

https://www.desmos.com/calculator/liejprtfic

Aaghaz Mahajan - 2 years, 8 months ago

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Thank you! I think you mean a2+b2+c2+6((ab)2+(bc)2+(ca)2)3(a4+b4+c4)2\sqrt{\frac{a^2+b^2+c^2+\sqrt{6((ab)^2+(bc)^2+(ca)^2)-3(a^4+b^4+c^4)}}2} when none of the interior angles is bigger than 120120^\circ

If AA is the area of the triangle, then it equals a2+b2+c2+43A2\sqrt{\frac{a^2+b^2+c^2+4\sqrt3A}2}

X X - 2 years, 8 months ago

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Yup.....that is what I have sent in the link!!!! :)

Aaghaz Mahajan - 2 years, 8 months ago

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@Aaghaz Mahajan Thanks!

X X - 2 years, 8 months ago

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@X X @X X No problemmo........anyways, I was thinking, a good follow up to the question would be..........What are the conditions on a b and c such that the given expression is an integer!!! For instance, a triangle with side lengths 3,5 and 7 satisfy the aforementioned property!!!!

Aaghaz Mahajan - 2 years, 8 months ago

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@Aaghaz Mahajan Interesting problem! Let's think about it.

X X - 2 years, 8 months ago

@Aaghaz Mahajan I looked for Fermat Point and found out also if one of the triangle's angle(ABC\angle ABC) is bigger than 120 degrees, then the point is on BB.

The triangle with side length 3,5,7 has one angle equal to 120 degrees, so this is probably the reason that it is an integer.

(So a triangle with integer lengths, and an angle not smaller than 120 degrees will satisfiy the aforementioned property. But if there is no angle bigger than 120 degrees?)

X X - 2 years, 8 months ago
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